Confused about waves

  • Thread starter SnowOwl18
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  • #1
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Ok, I understand waves and all, but for some reason I can't figure out how to manipulate the equations I know to solve these simple problems.

1. A standing sound wave in a pipe of length 2.70m is open at one end. Find the frequency of this harmonic. Use 342m/s for the speed of sound in air.

2. [a standing transverse wave on a flexible string] The string is 1.13m long with tension 10N. The total mass of the string is 8.80g. Find the period of the oscillation.

3. A typical steel B-string in a guitar resonates in its fundamental frequency at 240 Hertz. The length of the string is 0.620 m. What is the wave velocity along the string?

So for number 1 I tried to use the equation f=v/wavelength . But I think I can't just use the pipe length as the wavelenth because it's not the same. But I can't seem to figure out what to do about that.
For #2 I tried to use v= sqrt F/ (m/L)...but how do I get the period after solving for the velocity? I know I could use v=wavelength/T , but I don't know the wavelength...
And #3, same idea. Basically for all of them I have trouble with the wavelength. Can anyone help me figure out what concept is escaping me right now? Thanks so much.
 

Answers and Replies

  • #2
Pyrrhus
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Assuming for 1 and 2 fundamental frequency.

1) [tex] v = f \lambda [/tex]

Remember when one end's closed [itex] \lambda_{1} = 4L [/tex]

2) [tex] v = \sqrt{\frac{T}{\mu}} [/tex]

[tex] v = \sqrt{\frac{T}{\frac{m}{L}}} [/tex]

[tex] f = \frac{v}{\lambda} [/tex]

[tex] \lambda_{1} = 2L [/tex]

[tex] f = \frac{1}{T} [/tex]

3) [tex] \lambda_{1} = 2L [/tex]

[tex] v = f \lambda [/tex]
 
  • #3
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Oh thank you for all of the equations...those look like helpful ways to solve them. I just tried #1 assuming wavelength= 4L ...so I did 2.7 x 4 = 10.8 . 342m/s / 10.8m = 31.6 Hz...but the computer said I was wrong. How is that so?
 
  • #4
Pyrrhus
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Well for he first problem i don't know which harmonic they are refering,

[tex] f_{n} = \frac{1}{2L} v [/tex]

They only give us the speed and lenght of the pipe and ask us to find the harmonic, i don't think that's enough info, do they give us more? which harmonic?
 
  • #5
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nope...and i just tried that equation but it didn't work. i also tried what you gave for number 2 as well and that didn't work either...very peculiar, since they are the right equations and all. both questions have pictures that go with them, but they just show waves...i pasted the written problem and figured it didn't matter. i'm unsure of how to post the pictures...
 
  • #6
Pyrrhus
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The picture must be important just tell me how many nodes there are, Nodes are the points that remain unaffected (not on the parabolic arches). They look like intersections with the horizontal axis
 
  • #7
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there are 4 nodes in problem 1 and 5 in problem 2....but those don't include where the line begins on the horizontal axis..not sure if that part would count as a node.
 
  • #8
Pyrrhus
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SnowOwl18 said:
there are 4 nodes in problem 1 and 5 in problem 2....but those don't include where the line begins on the horizontal axis..not sure if that part would count as a node.

of course because there's one open end, that must be the 7th harmonic for the first problem (1 open end) n=7

[tex] f_{n} = n \frac{v}{4L} [/tex] where n=1,3,5,7 ...


The other problem is the 4th harmonic for the string (n=4).

[tex] f_{n} = \frac{n}{2L} \sqrt{\frac{T}{\mu}} [/tex]

where n = 1,2,3,...
 
Last edited:
  • #9
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I don't understand how number 1 works either. I have a problem exactly like it and I can't get it to work.
 
Last edited:

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